
#1
Oct1207, 06:45 AM

P: 167

A ball is thrown toward a cliff of height h with a speed of 30m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 4.0s later.
How high is the cliff? do you just use the equation h=.5gt^2 ? and put it in 9.80=g and 4=t ? 



#2
Oct1207, 07:31 AM

Mentor
P: 40,878

You also need to consider the initial speed in the ydirection; here's the complete equation:
[tex]y = y_0 + v_{0y} t  (1/2) g t^2[/tex] 



#3
Oct1207, 07:36 AM

P: 167

would this be correct: y0=26 , v0=30, g=9.8 , t=4 ... so y=26+(30x4) .5(9.8)(4)^2 = 224.4 m ?
i got vy by doing v0sintheta= 26 



#4
Oct1207, 07:44 AM

Mentor
P: 40,878

components of velocityy0 is your initial position, which I presume is on the ground at height = 0 v0 (which I'll change to v0y) is the vertical component of the initial velocity, what you call v0sin(theta) g = 9.8 m/s^2 To make it less confusing, I'll relabel v0 to be v0y in my equation. 



#5
Oct1207, 07:48 AM

P: 167

ohh sorry, so y=0 + 26(4)  .5(9.8)(4)^2 = 25.6 m




#7
Oct1207, 07:59 AM

P: 167

thanks! can you help me with the next part too..! What was the maximum height of the ball?
i think the equation is just the same as the one you wrote, but just cut t=4 in half to get the peak height so t=2 ..so y= 26(2).5(9.8)(2)^2 = 32.4 m ? 



#8
Oct1207, 08:10 AM

Mentor
P: 40,878

No, you can't assume that the peak is at half the time. After all, it lands up on a cliff, so it spent more time rising than falling. (If it fell back down to the original height, then you'd be correct.)
Instead, use a velocity equation for the ydirection to figure out the time when it reaches maximum height. Hint: At the maximum height, what's the vertical speed? 



#9
Oct1207, 08:14 AM

P: 167

um i think the vertical speed should be 0




#10
Oct1207, 08:16 AM

P: 167

so would it be okay to use the equation vfy= viy +ayT so 0= 26+9.8t so t= 2.65 and then cut then stick that time in that first equation you gave me?




#11
Oct1207, 08:18 AM

P: 167

so y= 26(2.65) .5(9.8)(2.65)^2 =34 .4m ?




#12
Oct1207, 08:31 AM

P: 167

are the equations i used alright?




#14
Oct1207, 08:39 AM

P: 167

thank you Doc Al! do you mind one more question, it will be the last one i promise! =)




#15
Oct1207, 08:41 AM

P: 167

What is the ball's impact speed?
i thought what you should do is find the final velocity in the x component and y component. and i thought that vfx= 15 since there is no acc in the x direction and vfy= 26 b/c of neg acc. and then i thought you should take the magnitude, but this does not work out to be right. do you know what i am doing wrong wait i think i did this wrong, look at my next post please 



#16
Oct1207, 08:45 AM

P: 167

oh no actually should vfy= 65.2 because i tried the equation vfy=viy+ay x T so 26+(9.8x4)= 65.2
so do i do square root (15^2 + 65.2^2) = 66.9 ? 



#18
Oct1207, 08:58 AM

P: 167

ohhh so should it be vfy= 13.2 and then do the sqaure root (13.2^2 +15^2 ) =19.98



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